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Balance the following redox reaction by the ion-electron method for both acidic and basic solutions.

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Balance the following redox reaction by the ion-electron method for both acidic and basic solutions.

Post  NeoTrion on Sun May 02, 2010 12:56 am

Having a problem solving a question here. This one has two equations. The second one I believe I have solved, but the first one is giving me some trouble. I can get what I think is the half reactions on the first one, but nothing as far as the acidic and basic solutions go...

Balance the following redox reaction by the ion-electron method for both acidic and basic solutions. Be sure to identify and write the half reactions for oxidation and reduction.

a. S2O3- (aq) + I2(aq) → S4O62- (aq) + I- (aq)
b. MnO4- (aq) + Fe2+ (aq) → Fe3+ (aq) + Mn2+ (aq)

a. S2O3- (aq) + I2(aq) → S4O62- (aq) + I- (aq)

2e- + I2 → 2I- Reduction
2S2O3- → S4O62- Oxidation

But if I am right there, then I am stuck and have nothing to add water to to make an acidic solution. Here is what I did on B.

b. MnO4- (aq) + Fe2+ (aq) → Fe3+ (aq) + Mn2+ (aq)

Fe2+ → Fe3+ + e- Oxidation
5e- + MnO4- → Mn+2 Reduction

5Fe2+ → 5Fe3+ + 5e-
8H+ + MnO4- + 5e- → Mn+2 + 4H20

8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O Acidic

8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
8OH- + 8H+ + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O + 8OH-

4H20 + MnO4- + 5Fe2+ → Mn2+ + 5Fe3+ + 8OH-

Hopefully someone can point me in the direction on A and check B to make sure I know what I am doing.

NeoTrion

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